જો int frac{cos x+x}{1+sin x} d x=f(x)+int fr | vedclass

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Question

(A) આપેલ છે કે $\int \frac{\cos x+x}{1+\sin x} d x=f(x)+\int \frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x+c_r$. $

Vedclass · Accepted Answer

$\frac{-2 x}{1+	an \frac{x}{2}}$

Vedclass · Answer

(A) આપેલ છે કે $\int \frac{\cos x+x}{1+\sin x} d x=f(x)+\int \frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x+c_r$. $f(x) = \int \left( \frac{\cos x+x}{1+\sin x} - \frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} ight) d x$. $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ અને $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ નો ઉપયોગ કરીને, આપણે સંકલ્યનું સાદું રૂપ આપીએ છીએ. નોંધો કે $\frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} = \frac{(3 \cos \frac{x}{2}-\sin \frac{x}{2})(\cos \frac{x}{2}+\sin \frac{x}{2})}{1+\sin x} = \frac{3 \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} - \sin^2 \frac{x}{2}}{1+\sin x} = \frac{2 \cos^2 \frac{x}{2} + \sin x + 1}{1+\sin x} = \frac{1+\cos x + \sin x}{1+\sin x} = \frac{\cos x}{1+\sin x} + 1$. આ કિંમત મૂકતા, $f(x) = \int \left( \frac{\cos x+x}{1+\sin x} - (\frac{\cos x}{1+\sin x} + 1) ight) d x = \int (\frac{x}{1+\sin x} - 1) d x$. $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ નો ઉપયોગ કરીને, આપણને $f(x) = \frac{-2x}{1+	an \frac{x}{2}}$ મળે છે. તેથી, સાચો વિકલ્પ $A$ છે.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

જો $\int \frac{\cos x+x}{1+\sin x} d x=f(x)+\int \frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x+c_r$ હોય, તો $f(x)=$

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